3.452+5.672+7.892+⋯=∑n=2∞(2n−1).(2n)(2n+1)2=∑n=2∞Un\frac{3.4}{5^2}+\frac{5.6}{7^2}+\frac{7.8}{9^2}+\dots = \sum_{n=2}^\infin\frac{(2n-1).(2n)}{(2n+1)^2}= \sum_{n=2}^\infin U_n523.4+725.6+927.8+⋯=∑n=2∞(2n+1)2(2n−1).(2n)=∑n=2∞Un
Then we have
limn→∞Un=limn→∞(2n−1).(2n)(2n+1)2=limn→∞(2−1n).2(2+1n)2=2.24=44=1≠0\lim_{n\to \infin} U_n= \lim_{n\to\infin}\frac{(2n-1).(2n)}{(2n+1)^2} = \lim_{n\to\infin}\frac{(2-\frac{1}{n}).2}{(2+\frac{1}{n})^2}= \frac{2.2}{4}=\frac{4}{4}=1 \ne0limn→∞Un=limn→∞(2n+1)2(2n−1).(2n)=limn→∞(2+n1)2(2−n1).2=42.2=44=1=0
Since limn→∞Un≠0\lim_{n\to \infin}U_n\ne0limn→∞Un=0 then by Cauchy's test for divergence the series is divergent(not convergent)
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